Quantitative Aptitude

Probability

1,860 Questions
All Sample Space Setup 558 Permutation & Combination Counting 288 Multiplication Theorem (AND) 234 Addition Theorem (OR) 151 At Least One Probability 134 Complement Rule (P(Not A)) 130 Drawing Without Replacement 126 Conditional Probability 75 Bayes' Theorem 69 Drawing With Replacement 49 Success and Failure Rates 46

Probability Questions

Multiple choice

A fair die is rolled repeatedly until a six is obtained. Let X denote the number of rolls required. The conditional probability that $X \geq 6$ given $X > 3$ equals
  1. $\displaystyle \frac{125}{216}$
  2. $\displaystyle \frac{25}{216}$
  3. $\displaystyle \frac{5}{36}$
  4. $\displaystyle \frac{25}{36}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a geometric distribution. P(X >= 6 | X > 3) = P(X >= 6) / P(X > 3). P(X > 3) = (5/6)^3. P(X >= 6) = (5/6)^5. Ratio = (5/6)^5 / (5/6)^3 = (5/6)^2 = 25/36.

Multiple choice

A bag contains $13$ balls numbered from $1$ to $13$. Suppose drawing of an even number is a sucess. Two balls are drawn with replacement from the bag. The probability of getting two successes is
  1. $\dfrac{84}{169}$
  2. $\dfrac{49}{169}$
  3. $\dfrac{36}{169}$
  4. $\dfrac{120}{169}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Even numbers in 1 to 13 are 2, 4, 6, 8, 10, 12 (6 numbers). Probability of success = 6/13. Since draws are with replacement, probability of two successes = (6/13) * (6/13) = 36/169.

Multiple choice

The probability of obtaining more tails than heads in $6$ tosses of a fair coins is?
  1. $2/64$
  2. $22/64$
  3. $21/64$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In 6 tosses, the total outcomes are 2^6 = 64. The probability of more tails than heads is the same as the probability of more heads than tails, and since the probability of equal heads and tails (3 each) is (6C3)/64 = 20/64, the remaining 44 outcomes are split equally between more heads and more tails, giving 22/64 each.

Multiple choice

A pack of cards consists of $15$ cards numbered $1$ to $15$. Three cards are drawn at random with replacement. Then, the probability of getting two odd and one even numbered cards, is?
  1. $\dfrac{348}{1125}$
  2. $\dfrac{398}{1125}$
  3. $\dfrac{448}{1125}$
  4. $\dfrac{498}{1125}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Odd numbers are {1, 3, 5, 7, 9, 11, 13, 15} (8 cards). Even numbers are {2, 4, 6, 8, 10, 12, 14} (7 cards). P(Odd) = 8/15, P(Even) = 7/15. We need 2 odd and 1 even in 3 draws. This is a binomial scenario: 3C2 * (8/15)^2 * (7/15)^1 = 3 * (64/225) * (7/15) = 1344 / 3375. Dividing by 3 gives 448 / 1125.

Multiple choice

A fair coin is tossed repeatedly. If tail appears on first four tosses then find the probability of head appearing on fifth toss
  1. $\displaystyle\frac{1}{2}$
  2. $\displaystyle\frac{1}{4}$
  3. $\displaystyle\frac{3}{4}$
  4. $\displaystyle\frac{2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Coin tosses are independent events. The outcome of the previous tosses does not affect the probability of the next toss. Therefore, the probability of getting a head on the fifth toss remains 1/2.

Multiple choice

The chance of throwing sum of $3$ or $5$ or $11$ with a pair of die is
  1. $ \displaystyle \frac{2}{9} $
  2. $ \displaystyle \frac{5}{9} $
  3. $ \displaystyle \frac{1}{9} $
  4. $ \displaystyle \frac{5}{36} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total outcomes for two dice = 36. Sum of 3: (1,2), (2,1) = 2 outcomes. Sum of 5: (1,4), (2,3), (3,2), (4,1) = 4 outcomes. Sum of 11: (5,6), (6,5) = 2 outcomes. Total favorable = 2 + 4 + 2 = 8. Probability = 8/36 = 2/9.

Multiple choice

The probability of getting at least one head when we toss $5$ unbiased coin is
  1. $\dfrac{5}{32}$
  2. $\dfrac{1}{32}$
  3. $\dfrac{31}{32}$
  4. $\dfrac{4}{32}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The probability of getting at least one head is 1 minus the probability of getting no heads. For 5 coins, the probability of no heads (all tails) is (1/2)^5 = 1/32. Thus, the probability of at least one head is 1 - 1/32 = 31/32.